Question

Find the vector equation of the plane which is at a distance of 7 units from the origin and normal to the vector $3\hat{i}+5\hat{j}-6\hat{k}$.

Answer 1

The normal vector is $\overrightarrow{n}=3\hat{i}+5\hat{j}-6\hat{k}$

Therefore,

$\hat{n}=\frac{\overrightarrow{n}}{|\overrightarrow{n}|}=\frac{3\hat{i}+5\hat{j}-6\hat{k}}{\sqrt{3^2+5^2+6^2}}=\frac{3\hat{i}+5\hat{j}-6\hat{k}}{\sqrt{70}}$

Equation of the plane with postion vector $\overrightarrow{n}$ is given by, $\overrightarrow{r}.\hat{n}=d$

$\hat{r}.\left(\frac{3\hat{i}+5\hat{j}-6\hat{k}}{\sqrt{70}}\right)=7$

This is the vector equation of the required plane.